In a close-packed structure of mixed oxides,t | vedclass

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(D) Let the number of $O^{2-}$ ions in the crystal lattice be $N$.Then,the number of octahedral voids $= N$.The number of tetrahedral voids $= 2N$.Num

Vedclass · Accepted Answer

$AB_2O_4$

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(D) Let the number of $O^{2-}$ ions in the crystal lattice be $N$.Then,the number of octahedral voids $= N$.The number of tetrahedral voids $= 2N$.Number of $A^{2+}$ ions occupying $1/8$ of tetrahedral voids $= \frac{1}{8} 	imes 2N = \frac{1}{4}N$.Number of $B^{3+}$ ions occupying $1/2$ of octahedral voids $= \frac{1}{2} 	imes N = \frac{1}{2}N$.The ratio of $A^{2+} : B^{3+} : O^{2-}$ is $\frac{1}{4}N : \frac{1}{2}N : N$.Multiplying by $4$,we get the ratio $1 : 2 : 4$.Therefore,the formula of the mixed oxide is $AB_2O_4$.

In a close-packed structure of mixed oxides,the lattice is composed of oxide ions. One-eighth of the tetrahedral voids are occupied by divalent cations $(A^{2+})$,while one-half of the octahedral voids are occupied by trivalent cations $(B^{3+})$. The formula of the mixed oxide is:

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